CS6515 ALGORITHMS ACTUAL EXAM 1 PAPER
2026 FULL QUESTIONS WITH VERIFIED
ANSWERS
◉ DP: Types of Subproblems (4). Answer: Input = x1, x2, ..., xn
1) Subproblem = x1, x2, ..., xi ; O(n)
2) Subproblem = xi, xi+1, ..., xj ; O(n^2)
Input = x1, x2, ..., xn; y1, y2, ..., ym
1) Subproblem = x1, x2, ..., xi; y1, y2, ..., yj ; O(mn)
Input = Rooted Binary Tree
1) Subproblem = Smaller rooted binary tree inside the Input.
◉ DC: Geometric Series. Answer: Given r = common ratio and a =
first term in series
=> a + ar + ar^2 + ar^3 + ... + ar^(n-1)
=> a * [(1 - r^n) / (1-r)]
,◉ DC: Arithmetic Series. Answer: Given d = common difference and
a = first term in series => a + (a + d) + (a + 2d) + ... + (a + (n-1)d
Sum = n/2 * [2*a + (n-1)d]
◉ DC: Solving Recurrences - Master Theorem. Answer: If T(n) =
aT([n/b]) + O(n^d) for constants a>0, b>1, d>=0:
T(n) = {
O(n^d) if d > logb(a)
O((n^d)logn) if d = logb(a)
O(n^(logb(a))) if d < logb(a)
}
◉ Nth roots of Unity. Answer: (1, 2*PI*j/n) for j = 0, 1, ..., n-1
*Around the Unit Circle!
◉ Steps to solve for FFT. Answer: 1) Write out Matrix Coefficient
Form based on n (size of input) Mn(w) = [ 1 1 ... 1
1 w ... w^n-1
...
1 w^n-1 ... w^((n-1)*(n-1)) ]
, 2) Find value for w = e^(2*PI*i)/n, Substitute in Mn(w).
3) For the input coefficients into nx1 matrix. I.E. [4 0 1 1], let known
as B.
4) Evaluate FFT:
a) FFT of Input = Mn(w) x B
b) Inverse FFT of Input = 1/n * Mn(w^-1) x B
◉ Euler's Formula. Answer: e^ix = cosx + isinx
◉ Imaginary Number Multiples. Answer: i = i, i^2 = -1, i^3 = -i, i^4 =
1
i = -i, i^2 = -1, i^3 = i, i^4 = 1
◉ Omega(w). Answer: w = (1, 2*PI / n) = e^(2*PI*i/n)
◉ DC Algorithms and Runtimes (6). Answer: MergeSort: O(nlogn) -
Split the input array into two halves and recursively sort and merge
at the end.
QuickSort: O(nlogn) - Same splitting strategy as MergeSort.
2026 FULL QUESTIONS WITH VERIFIED
ANSWERS
◉ DP: Types of Subproblems (4). Answer: Input = x1, x2, ..., xn
1) Subproblem = x1, x2, ..., xi ; O(n)
2) Subproblem = xi, xi+1, ..., xj ; O(n^2)
Input = x1, x2, ..., xn; y1, y2, ..., ym
1) Subproblem = x1, x2, ..., xi; y1, y2, ..., yj ; O(mn)
Input = Rooted Binary Tree
1) Subproblem = Smaller rooted binary tree inside the Input.
◉ DC: Geometric Series. Answer: Given r = common ratio and a =
first term in series
=> a + ar + ar^2 + ar^3 + ... + ar^(n-1)
=> a * [(1 - r^n) / (1-r)]
,◉ DC: Arithmetic Series. Answer: Given d = common difference and
a = first term in series => a + (a + d) + (a + 2d) + ... + (a + (n-1)d
Sum = n/2 * [2*a + (n-1)d]
◉ DC: Solving Recurrences - Master Theorem. Answer: If T(n) =
aT([n/b]) + O(n^d) for constants a>0, b>1, d>=0:
T(n) = {
O(n^d) if d > logb(a)
O((n^d)logn) if d = logb(a)
O(n^(logb(a))) if d < logb(a)
}
◉ Nth roots of Unity. Answer: (1, 2*PI*j/n) for j = 0, 1, ..., n-1
*Around the Unit Circle!
◉ Steps to solve for FFT. Answer: 1) Write out Matrix Coefficient
Form based on n (size of input) Mn(w) = [ 1 1 ... 1
1 w ... w^n-1
...
1 w^n-1 ... w^((n-1)*(n-1)) ]
, 2) Find value for w = e^(2*PI*i)/n, Substitute in Mn(w).
3) For the input coefficients into nx1 matrix. I.E. [4 0 1 1], let known
as B.
4) Evaluate FFT:
a) FFT of Input = Mn(w) x B
b) Inverse FFT of Input = 1/n * Mn(w^-1) x B
◉ Euler's Formula. Answer: e^ix = cosx + isinx
◉ Imaginary Number Multiples. Answer: i = i, i^2 = -1, i^3 = -i, i^4 =
1
i = -i, i^2 = -1, i^3 = i, i^4 = 1
◉ Omega(w). Answer: w = (1, 2*PI / n) = e^(2*PI*i/n)
◉ DC Algorithms and Runtimes (6). Answer: MergeSort: O(nlogn) -
Split the input array into two halves and recursively sort and merge
at the end.
QuickSort: O(nlogn) - Same splitting strategy as MergeSort.
